Question 112446
The way you go about this depends on whether you are studying calculus and know how to take the first and second derivatives of a polynomial function or not.  Not knowing, I'll do it two different ways:


First the calculus way.

The first derivative gives us the slope of a line tangent to the curve at any point in the domain of the function.  Since the tangent line at a local minimum or maximum on the curve is horizontal, and the slope of a horizontal line is zero, you simply set the first derivative equal to zero and solve to find the x-coordinate of the extrema.  In this case, given:


{{{g(x)=-4x^2+8x+3}}}
{{{dg(x)/dx=-8x+8}}}
{{{-8x+8=0}}}
{{{-8x=-8}}}
{{{x=1}}}


Knowing the x-coordinate of the point allows us to find the y-coordinate by evaluating g(1).


{{{g(1)=-4(1)^2+8(1)+3=-4+8+3=7}}}


And the critical point is the point (1, 7)


Now we still have to determine whether the extreme point is a minimum or a maximum.  If it is a maximum, the slope of a tangent to the curve will change from positive to negative as x moves left to right across the critical point.  Conversely, if it is a minimum, the tangent slope will change from negative to positive as you move across.  The second derivative, being the rate of change of the rate of change, evaluated at the x-coordinate of the critical point will tell us by its sign which way it is going.  If the second derivative at the point is negative, then we are at a maximum.  If the second derivative is positive, it is a minimum.


{{{g(x)=-4x^2+8x+3}}}
{{{dg(x)/dx=-8x+8}}}
{{{d^2g(x)/dx^2=-8}}}


Since the second derivative is negative, this is a local maximum.



Now for the non-calculus way.  Since we are dealing with a 2nd degree polynomial, we know that if we graphed it, we would see a parabola.  The vertex (h, k) of a parabola {{{f(x)=ax^2+bx+c}}} is determined by the following:


{{{h=(-b)/2a}}}, and
{{{k=f(h)=ah^2+bh+c}}}


In our case:
{{{h=(-8)/2(-4)=1}}}, and
{{{k=f(1)=-4(1)^2+8(1)+3=7}}}


So the vertex, which is the critical point, is at (1, 7)


Since the coefficient on the {{{x^2}}} term is negative, i.e. -4, we know that the parabola is concave down.  That means the vertex is at the top and it represents a maximum.


Let's see if all this makes sense graphically:
{{{graph(400,400,-3,3,-2,8,-4x^2+8x+3,7)}}}


Looks good to me.  How about you?



Hope this helps,
John